Exponents Class- 7th RS Aggarwal Exe-5 A Goyal Brothers ICSE Math Solution . We provide step by step Solutions of lesson-5 Exponents for ICSE Class-7 Foundation RS Aggarwal Mathematics of Goyal Brothers Prakashan . Our Solutions contain all type Questions of Exe-5 A to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics. laws of exponents pdf class 7 / 6 with examples and worksheet.
Exponents Class- 7th RS Aggarwal Exe-5 A Goyal Brothers ICSE Math Solution
| Board | ICSE |
| Publications | Goyal brothers Prakashan |
| Subject | Maths |
| Class | 7th |
| Chapter-5 | Exponents |
| Writer | RS Aggrawal |
| Book Name | Foundation |
| Topics | Solution of Exe-5 A |
| Academic Session | 2023 – 2024 |
Exercise – 5 A
Exponents Class- 7th RS Aggarwal Exe-5 A Goyal Brothers ICSE Math Solution
1. Evaluate :
(i) 74
Solution: 74
= 7 × 7 × 7 × 7
= 2401
(ii) (-5)3
Solution: (-5)3
= (-5) × (-5) × (-5)
= -125
(iii) (3/4)5
Solution: (3/4)5
= (3/4) × (3/4) × (3/4) × (3/4) × (3/4)
= 243/1024
(iv) (-5/2)2
Solution: (-5/2)2
= (-5/2) × (-5/2)
= (25/4)
= 6(1/4)
2. Write as a power of 10 :
(i) 10000
Solution: 10000
= 104
(ii) One crore
Solution: One crore
= 107
(iii) One million
Solution: One million
= 106
3. Express each of the following in exponential notation :
(i) (-7/13) × (-7/13) × (-7/13)
Solution: (-7/13) × (-7/13) × (-7/13)
= (-7/13)3
(ii) (-8/3) × (-8/3) × (-8/3) × (-8/3)
Solution: (-8/3) × (-8/3) × (-8/3) × (-8/3)
= (-8/3)4
4. Express each of the following in exponential notation :
(i) (343/512)
Solution: (343/512)
= (7/8) × (7/8) × (7/8)
= (7/8)3
(ii) (-32/243)
Solution: (-32/243)
= (-2/3) × (-2/3) × (-2/3) × (-2/3) × (-2/3)
= (-2/3)5
(iii) (-1/128)
Solution: (-1/128)
= (-1/2) × (-1/2) × (-1/2) × (-1/2) × (-1/2) × (-1/2) × (-1/2)
= (-1/2)7
(iv) (729/64)
Solution: (729/64)
= (3/2) × (3/2) × (3/2) × (3/2) ×(3/2) × (3/2)
= (3/2)6
5. Express each of the following in exponential notation :
(i) (5/21)3 × (5/21)8
Solution: (5/21)3 × (5/21)8
= (5/21)3 + 8
= (5/21)11
(ii) (-7/3)11 × (-7/3)13
Solution: (-7/3)11 × (-7/3)13
= (-7/3)11 + 13
= (-7/3)24
(iii) (13/43)7 ÷ (13/43)2
Solution: (13/43)7 ÷ (13/43)2
= (13/43)7 – 2
= (13/43)5
(iv) (-16/35)16 ÷ (-16/35)3
Solution: (-16/35)16 ÷ (-16/35)3
= (-16/35)16 – 3
= (-16/35)13
(v) (-7/15)12 ÷ (-7/15)15
Solution: (-7/15)12 ÷ (-7/15)15
= (-7/15)12 – 15
= (-7/15)-3
= (-15/7)3
(vi) (1/24)13 ÷ (1/24)16
Solution: (1/24)13 ÷ (1/24)16
= (1/24)13 – 16
= (1/24)3
6. Simplify and express each of the following as a rational number :
(i) (6/5)3 × (5/2)2
Solution: (6/5)3 × (5/2)2
= (6× 6 × 6)/ (5 × 5 × 5) × (5 × 5) / (2 × 2)
= (9 × 6)/5
= (54/5)
= 10(4/5)
(ii) (3/4)2× (-1/2)5 × 23
Solution: (3/4)2× (-1/2)5 × 23
= (3 × 3)/(4 × 4) × (-1 × -1 × -1 × -1 × -1)/(2 × 2 × 2 × 2 × 2) × 2 × 2 × 2
=
=
= (-9/64)
(iii) (5/4)2× (2/3)2 × (-3/5)3
Solution: (5/4)2× (2/3)2 × (-3/5)3
= (5 × 5)/(4 × 4) × (2 × 2)/(3 × 3) × (-3 × -3 × -3)/(5 × 5 × 5)
= (25/16) × (4/9) × (-27/125)
= (-3/20)
(iv) (-3/4)3× (-5/2)3 × (2/3)5
Solution: (-3/4)3× (-5/2)3 × (2/3)5
= (-3 × -3 × -3)/(4 × 4 × 4) × (-5 × -5 × -5)/(2 × 2 × 2) × (2 × 2 × 2 × 2 × 2)/(3 × 3 × 3 × 3 × 3)
= (-27/64) × (-125/8) × (32/243)
= (125/144)
(v) (7/11)6 ÷ (7/11)3
Solution: (7/11)6 ÷ (7/11)3
= (7/11)6-3
= (7/11)3
= (7/11) × (7/11) × (7/11)
= (343/1331)
(vi) (-4/3)8 ÷ (-4/3)12
Solution: (-4/3)8 ÷ (-4/3)12
= (3/-4)8-12
= (3/-4)4
= (3/-4) × (3/-4) × (3/-4) × (3/-4)
= (81/256)
7. Simplify and express each of the following as a rational number :
(i) 102 × 152 / 22 × 3 × 55 × 64
Solution: 102 × 152 / 22 × 3 × 55 × 64
= (100 × 3375)/(4 × 3 × 3125 × 1296) 1296 × 6
= (1/144)
(ii) 35 × 25 × 105 / 57 × 65
Solution: 35 × 25 × 105 / 57 × 65
= (243 × 25 × 100000)/(78125 × 7776)
= 1
8. The distance between the Earth and the Moon is approximately 384000 km. Express this distance in meters in exponential notation.
Solution: We know that 1 km = 1000m
= 103
= 384000 km = (384000 × 1000)m
= (384 × 1000000)m
= 384 × 106m
9. The RAM of a computer is 8 gigabyte. If each gigabyte is equal to 109 bytes, then express the RAM in bytes.
Solution: 1 gigabyte = 109 bytes
8 gigabyte = 8 × 109 bytes
= 8000000000bytes
10. In a tennis competition, 128 players were selected for a series of knockout rounds. In each round the losers were eliminated and the winners reached the next round. How many players moved to the next round after 4th round? Express this number in the exponential notation in terms of the initial number of players.
[Hint. Required no of players = 128/24]
Solution: Number of players = 128
Number of matches = 128/2
4th round = 128/24
11. Express the following in centimeters (cm) in exponential notation :
(i) 98 hm
Solution: 98 hm
= We know :
1 hm = 10 dam
1 dam = 10 m
1 m = 10 dm
1 dm = 10 cm
98 hm
= 98 × 10 × 10 × 10 × 10
= 98 × 104cm
= 980000 cm
(ii) 157 km
Solution: 157 km
= We know :
1 km = 10 hm
1 hm = 10 dam
1 dam = 10 m
1 m = 10 dm
1 dm = 10 cm
157 km
= 157 × 10 × 10 × 10 × 10 × 10
= 157 × 105 km
= 15700000 km
(iii) 371 m
Solution: 371 m
1 m = 10 dm
1 dm = 10 cm
371 m
= 371 × 10 × 10
= 371 × 102cm
= 37100cm
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